Question

Problem 5: Given: 1.0 M HOAc solution (K4 = 1.8 x 10-5). 1.0 M NaOAc solution Make 500 mL of 1.0 M HOAc/OAc- buffer with pH = 5.00

Answer 1

Ans:

pKa of acetic acid (HOAc) = -logKa = -log(1.8 x 10^-5) =4.74

pH of an acid buffer can be calculated using Henderson–Hasselbalch equation;

pH = pKa + log ([Conjugate base]/[acid]) = pKa + log ([OAc^-]/[HOAc])

pH = pKa + log ([NaOAc]/[HOAc])

5 = 4.74 + log ([NaOAc]/[HOAc])

Required buffer component ratio = [NaOAc]/[HOAc] = 1.8197

Let ‘n1’ be the no. of moles of NaOAc & ‘n2’ be the no. of moles of HOAc required to make the buffer.

Volume of buffer required, V = 500 mL = 0.5 L

[NaOAc]/[HOAc] = (n1/V)/(n2/V) = 1.8197                 {Molarity = no of moles/volume}

n1 = 1.8197n2

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Let V1 & V2 be the volume of NaOAc (1.0 M = M1) & HOAc (1.0 M = M2) taken respectively to form the buffer.

Volume of buffer required, V = 500 mL = 0.5 L

V1 + V2 = V

(n1/M1) + (n2/M2) = V

(1.8197n2/1.0) + (n2/1.0) = 0.5 L

n2 = 0.177 mol

n1 = 1.8197n2 = 0.323 mol

V1 = n1/M1 = 0.322/1.0 = 0.323 L = 323 mL

V2 = n2/M2 = 0.177/1.0 = 0.177 L = 177 mL

323 mL of 1.0 M NaOAc mixed with 177 mL of 1.0 M HOAc gives us 500 mL of buffer with pH 5